Lemma 106.10.2. Let $f : \mathcal{X} \to Y$ be a morphism from an algebraic stack to an algebraic space. Let $V \subset Y$ be an open subspace. Assume

1. $f$ is separated and of finite type,

2. $Y$ is quasi-compact and quasi-separated,

3. $V$ is quasi-compact, and

4. $\mathcal{X}_ V$ is a gerbe over $V$.

Then there exists a commutative diagram

$\xymatrix{ \overline{Z} \ar[rd]_{\overline{g}} & Z \ar[l]^ j \ar[d]_ g \ar[r]_ h & \mathcal{X} \ar[ld]^ f \\ & Y }$

with $j$ an open immersion, $\overline{g}$ and $h$ proper, and such that $|V|$ is contained in the image of $|g|$.

Proof. Suppose we have a commutative diagram

$\xymatrix{ \mathcal{X}' \ar[d]_{f'} \ar[r] & \mathcal{X} \ar[d]^ f \\ Y' \ar[r] & Y }$

and a quasi-compact open $V' \subset Y'$, such that $Y' \to Y$ is a proper morphism of algebraic spaces, $\mathcal{X}' \to \mathcal{X}$ is a proper morphism of algebraic stacks, $V' \subset Y'$ maps surjectively onto $V$, and $\mathcal{X}'_{V'}$ is a gerbe over $V'$. Then it suffices to prove the lemma for the pair $(f' : \mathcal{X}' \to Y', V')$. Some details omitted.

Overall strategy of the proof. We will reduce to the case where the image of $f$ is open and $f$ has a section over this open by repeatedly applying the above remark. Each step is straightforward, but there are quite a few of them which makes the proof a bit involved.

Using Limits of Spaces, Proposition 70.16.1 we reduce to the case where $Y$ is a scheme. (Let $Y' \to Y$ be a finite surjective morphism where $Y'$ is a scheme. Set $\mathcal{X}' = \mathcal{X}_{Y'}$ and apply the initial remark of the proof.)

Using Lemma 106.9.1 (and Morphisms of Stacks, Lemma 101.28.8 to see that a gerbe is flat and locally of finite presentation) we reduce to the case where $f$ is flat and of finite presentation.

Since $f$ is flat and locally of finite presentation, we see that the image of $|f|$ is an open $W \subset Y$. Since $\mathcal{X}$ is quasi-compact (as $f$ is of finite type and $Y$ is quasi-compact) we see that $W$ is quasi-compact. By Lemma 106.10.1 we can find a finite surjective morphism $g : Y' \to Y$ such that $g^{-1}(W) \to Y$ factors Zariski locally through $\mathcal{X} \to Y$. After replacing $Y$ by $Y'$ and $\mathcal{X}$ by $\mathcal{X} \times _ Y Y'$ we reduce to the situation described in the next paragraph.

Assume there exists $n \geq 0$, quasi-compact opens $W_ i \subset Y$, $i = 1, \ldots , n$, and morphisms $x_ i : W_ i \to \mathcal{X}$ such that (a) $f \circ x_ i = \text{id}_{W_ i}$, (b) $W = \bigcup _{i = 1, \ldots , n} W_ i$ contains $V$, and (c) $W$ is the image of $|f|$. We will use induction on $n$. The base case is $n = 0$: this implies $V = \emptyset$ and in this case we can take $\overline{Z} = \emptyset$. If $n > 0$, then for $i = 1, \ldots , n$ consider the reduced closed subschemes $Y_ i$ with underlying topological space $Y \setminus W_ i$. Consider the finite morphism

$Y' = Y \amalg \coprod \nolimits _{i = 1, \ldots , n} Y_ i \longrightarrow Y$

and the quasi-compact open

$V' = (W_1 \cap \ldots \cap W_ n \cap V) \amalg \coprod _{i = 1, \ldots , n} (V \cap Y_ i).$

By the initial remark of the proof, if we can prove the lemma for the pairs

$(\mathcal{X} \to Y, W_1 \cap \ldots \cap W_ n \cap V) \quad \text{and}\quad (\mathcal{X} \times _ Y Y_ i \to Y_ i, V \cap Y_ i),\quad i = 1, \ldots , n$

then the result is true. Here we use the settheoretic equality $V = (W_1 \cap \ldots \cap W_ n \cap V) \cup \bigcup \nolimits _{i = 1, \ldots n} (V \cap Y_ i)$. The induction hypothesis applies to the second type of pairs above. Hence we reduce to the situation described in the next paragraph.

Assume there exists $n \geq 0$, quasi-compact opens $W_ i \subset Y$, $i = 1, \ldots , n$, and morphisms $x_ i : W_ i \to \mathcal{X}$ such that (a) $f \circ x_ i = \text{id}_{W_ i}$, (b) $W = \bigcup _{i = 1, \ldots , n} W_ i$ contains $V$, (c) $W$ is the image of $|f|$, and (d) $V \subset W_1 \cap \ldots \cap W_ n$. The morphisms

$T_{ij} = \mathit{Isom}_\mathcal {X}(x_ i|_{W_ i \cap W_ j \cap V}, x_ j|_{W_ i \cap W_ j \cap V}) \longrightarrow W_ i \cap W_ j \cap V$

are surjective, flat, and locally of finite presentation (Morphisms of Stacks, Lemma 101.28.10). We apply Lemma 106.10.1 to each quasi-compact open $W_ i \cap W_ j \cap V$ and the morphisms $T_{ij} \to W_ i \cap W_ j \cap V$ to get finite surjective morphisms $Y'_{ij} \to Y$. After replacing $Y$ by the fibre product of all of the $Y'_{ij}$ over $Y$ we reduce to the situation described in the next paragraph.

Assume there exists $n \geq 0$, quasi-compact opens $W_ i \subset Y$, $i = 1, \ldots , n$, and morphisms $x_ i : W_ i \to \mathcal{X}$ such that (a) $f \circ x_ i = \text{id}_{W_ i}$, (b) $W = \bigcup _{i = 1, \ldots , n} W_ i$ contains $V$, (c) $W$ is the image of $|f|$, (d) $V \subset W_1 \cap \ldots \cap W_ n$, and (e) $x_ i$ and $x_ j$ are Zariski locally isomorphic over $W_ i \cap W_ j \cap V$. Let $y \in V$ be arbitrary. Suppose that we can find a quasi-compact open neighbourhood $y \in V_ y \subset V$ such that the lemma is true for the pair $(\mathcal{X} \to Y, V_ y)$, say with solution $\overline{Z}_ y, Z_ y, \overline{g}_ y, g_ y, h_ y$. Since $V$ is quasi-compact, we can find a finite number $y_1, \ldots , y_ m$ such that $V = V_{y_1} \cup \ldots \cup V_{y_ m}$. Then it follows that setting

$\overline{Z} = \coprod \overline{Z}_{y_ j},\quad Z = \coprod Z_{y_ j},\quad \overline{g} = \coprod \overline{g}_{y_ j},\quad g = \coprod g_{y_ j},\quad h = \coprod h_{y_ j}$

is a solution for the lemma. Given $y$ by condition (e) we can choose a quasi-compact open neighbourhood $y \in V_ y \subset V$ and isomorphisms $\varphi _ i : x_1|_{V_ y} \to x_ i|_{V_ y}$ for $i = 2, \ldots , n$. Set $\varphi _{ij} = \varphi _ j \circ \varphi _ i^{-1}$. This leads us to the situation described in the next paragraph.

Assume there exists $n \geq 0$, quasi-compact opens $W_ i \subset Y$, $i = 1, \ldots , n$, and morphisms $x_ i : W_ i \to \mathcal{X}$ such that (a) $f \circ x_ i = \text{id}_{W_ i}$, (b) $W = \bigcup _{i = 1, \ldots , n} W_ i$ contains $V$, (c) $W$ is the image of $|f|$, (d) $V \subset W_1 \cap \ldots \cap W_ n$, and (f) there are isomorphisms $\varphi _{ij} : x_ i|_ V \to x_ j|_ V$ satisfying $\varphi _{jk} \circ \varphi _{ij} = \varphi _{ik}$. The morphisms

$I_{ij} = \mathit{Isom}_\mathcal {X}(x_ i|_{W_ i \cap W_ j}, x_ j|_{W_ i \cap W_ j}) \longrightarrow W_ i \cap W_ j$

are proper because $f$ is separated (Morphisms of Stacks, Lemma 101.6.6). Observe that $\varphi _{ij}$ defines a section $V \to I_{ij}$ of $I_{ij} \to W_ i \cap W_ j$ over $V$. By More on Morphisms of Spaces, Lemma 76.39.6 we can find $V$-admissible blowups $p_{ij} : Y_{ij} \to Y$ such that $s_{ij}$ extends to $p_{ij}^{-1}(W_ i \cap W_ j)$. After replacing $Y$ by the fibre product of all the $Y_{ij}$ over $Y$ we get to the situation described in the next paragraph.

Assume there exists $n \geq 0$, quasi-compact opens $W_ i \subset Y$, $i = 1, \ldots , n$, and morphisms $x_ i : W_ i \to \mathcal{X}$ such that (a) $f \circ x_ i = \text{id}_{W_ i}$, (b) $W = \bigcup _{i = 1, \ldots , n} W_ i$ contains $V$, (c) $W$ is the image of $|f|$, (d) $V \subset W_1 \cap \ldots \cap W_ n$, and (g) there are isomorphisms $\varphi _{ij} : x_ i|_{W_ i \cap W_ j} \to x_ j|_{W_ i \cap W_ j}$ satisfying

$\varphi _{jk}|_ V \circ \varphi _{ij}|_ V = \varphi _{ik}|_ V.$

After replacing $Y$ by another $V$-admissible blowup if necessary we may assume that $V$ is dense and scheme theoretically dense in $Y$ and hence in any open subspace of $Y$ containing $V$. After such a replacement we conclude that

$\varphi _{jk}|_{W_ i \cap W_ j \cap W_ k} \circ \varphi _{ij}|_{W_ i \cap W_ j \cap W_ k} = \varphi _{ik}|_{W_ i \cap W_ j \cap W_ k}$

by appealing to Morphisms of Spaces, Lemma 67.17.8 and the fact that $I_{ik} \to W_ i \cap W_ j$ is proper (hence separated). Of course this means that $(x_ i, \varphi _{ij})$ is a desent datum and we obtain a morphism $x : W \to \mathcal{X}$ agreeing with $x_ i$ over $W_ i$ because $\mathcal{X}$ is a stack. Since $x$ is a section of the separated morphism $\mathcal{X} \to W$ we see that $x$ is proper (Morphisms of Stacks, Lemma 101.4.9). Thus the lemma now holds with $\overline{Z} = Y$, $Z = W$, $\overline{g} = \text{id}_ Y$, $g = \text{id}_ W$, $h = x$. $\square$

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